Answer to Question #190971 in General Chemistry for kinzang

Question

 Here is a balanced equation. Use it to answer the questions below.

2Pb(NO3)2     🠖     2PbO     +     4NO2    +     O2

(i). How many moles of nitrogen dioxide would you be able to get from 2 moles of lead(II) nitrate?

(ii). How many moles of oxygen would you be able to get from 1 mole of lead (II) nitrate? 

(iii). What mass of PbO could be obtained from 200g of Pb(NO3)2?

(iv). If 130.0g of PbO are actually obtained when the experiment is done in the laboratory, what is the percentage yield?

Answer 1

$2Pb(NO_3)_2 \rightarrow 2PbO+4NO_2+O_2$

a.) 2 moles of $Pb(NO_3)_2$ gives 2 moles of $NO_2$

b) 2 moles of $Pb(NO_3)_2$ gives 1mole of $O_2$

Hence, 1 mole of $Pb(NO_3)_2$ gives $\dfrac{1}{2} \hspace{2mm}mole \hspace{2mm}of\hspace{2mm} O_2.$

c.) 1 mole of $Pb(NO_3)_2$ i.e. $331.2g /mol$ gives 1 mole i.e. $223.2g/mol$

Hence, 200g of $Pb(NO_3)_2$ gives $= \dfrac{223.2}{331.2} \times 200 = 134.78g$

d.) % $Yield = \dfrac{Actual\hspace{2mm} yield}{Theoretical\hspace{2mm} yield} \times 100$ %